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assignment problem for DOE course…

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  • #27431

    Kiana desperate!
    Participant

    this question is from a book but the instructor didn’t specify which one :-/
    It goes kinda like this:
    the experimental results on the effect of temperature and fabric types are provided in the table below. Analyze the data for a significant effect associated with temperature, fabric, interaction. (4 levels of temperature and 3 types of fabric with two # of data for the test appearing in the table)
    Is this a 2 factor ANOVA by sum of squares with replication??
    Because that’s how I tried solving it but the sum of squares total is almost the same as the sum of squares for the interaction (264.77 & 264.12) and the sum for fabric is also 41.4 which doesn’t add up and the sum for error would be negative…
    Should I only calculate the total, the fabric and temperature and use the interaction as the error term by subtraction??
    I would really appreciate it any help I can get.

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    #67155

    Tierradentro
    Participant

    Use General Linear Modeling, specifying Temperature as a Covariate. Treating Temperature as a categorical variable is likely throwing you off.

    You could also use regression if you reduce the fabric type to two levels.

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    #67158

    “Ken”
    Participant

    John,

    I believe GLM could do the job, but why use temperature as a covariate? I would look at this as a mixed level design. Interested in your thoughts.

    Ken

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    #67161

    Raju
    Participant

    It seems like a full 4X3 factorial design with two replications. What is the response variabe? Is it continuos? If it is, then you can analyse using standard full factorial model up using ANOVA technique. It is a standard analysis you can see in any of the statistical software.

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    #67165

    Neil Polhemus
    Participant

    If the data is balanced, as it appears to be (2 replicates at each of the 12 factor combinations), then I suggest you check your arithmetic. The factor and interaction sums of squares must add to less than the total and the error sum of squares can’t be negative.

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    #67167

    Kiana
    Participant

    Unfortunately, none of the things you mention sound familiar to me.
    I guess we didn’t reach there yet with the studies,yet.
    Neil, I have checked the math over and over.
    At the risk of sounding too weird, I’m going to copy the whole data here coz I cannot, for the life of me, figure out what I did wrong!!
    temp levels——210C 215 220 225
    type fabric A 1.8&2.1 2.0&2.1 4.6&5.0 7.5&7.9
    B 2.2&2.4 4.2&4.0 5.4;5.6 9.8;9.2
    C 2.8;3.2 4.4;4.8 8.7;8.4 13.2;13.0
    And here’s what I did in terms of calculations
    I treated it as a hypothesis testing where Ho: there is no difference between the types of fabric @ the various
    levels of temp. Against the alternative. And went to work. For SST took every piece of data and squared it minus the total of the data squared and divided by 24.
    SSTotal= 1016.29 – 18036.49/24 = 264.77
    SStemp= 5766.15/4×2 – 18036.49/24 = 41.4 where 5766.15 is the total from adding each squared column divided by the number of columnsXreplicates.
    SSfabric= 6343.1/3×2 – 18036.49/24 = 209.5
    6343.1 being the summation of sqared totals from each row : by number of rows X replicates.
    For the interaction I added the two replicates, squared that total, added them up, divided by 2 for # of replicates and came up with
    SSInter’n= 2031.2/2 – 18036.49/24= 264.12 !!!
    Please Neil, tell me I’m not completely stupid!
    I actually took this course because of the math involved (it’s an elective as part of a certificate program). The more I look at it, the less likely I am to see what I did wrong. I appreciate your help!

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    #67170

    Stephen Halliday
    Member

    In calculating the interaction SS you need to subtract the Temp ss and the Fabric SS.
    Thus 264.12 – 209.50 – 41.40 = 13.22

    This will leave an error ss of 0.65

    Hope this helps.

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    #67232

    Tierradentro
    Participant

    Hi Ken,

    Sorry for the delayed response to your question – if you analyze Temperature as a covariate you end up with fewer degrees of freedom in the model, as well as the ability to explore Temp**2, and Temp**2 * Fabric.

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    #67233

    Kiana
    Participant

    Would you like to try the chat coz I have more questions… sunday, 1:35 pm

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